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3 years ago

PLEASE HELP!! I HAVE TO TURN THIS IN IN 10 MINUTES

Mathematics

1 answer:

grin007 [14]3 years ago

6 0

Answer:

0.8

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer/how I got this answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

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A particular baseball player has a batting average (number of hits/number of times at bat) of 0.364. What is the probability tha

MAXImum [283]

The correct answer is option (c) P(not a hit) = 0.636

The batting average of the baseball player = $\frac{number of hits}{number of times at bat}$ = 0.364

Now, the probability of occurrence of any event is say, P, the probability of not occurrence of the event is given as = 1 - P

Therefore, the probability that he will not get a hit at his next at bat = 1- 0.364 = 0.636

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3 years ago

Two cards are selected from a standard deck of 52 playing cards. the first is replaced before the second card is selected. find

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The correct answer is
 Find the probability of selecting a diamond and then selecting a queen.

standard deck of 52 playing cards

hearts: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A---- > 13 cards

clubs: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A---- > 13 cards

spades: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A---- > 13 cards

diamonds: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A---- > 13 cards

a) The probability of selecting a diamond is 13/52

b) The probability of selecting a queen is 4/52

c) The probability of selecting a diamond and then selecting a queen is

(13/52)*(4/52)=0.0192------------->1.92%

6 0

3 years ago

The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr

aniked [119]

Answer:

(a) $E(X) = 0.383$

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) $P(x = 4) = 0.000169$

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$E(X) = \frac{n \times r}{N}$

$E(X) = \frac{10 \times 8}{209}$

$E(X) = \frac{80}{209}$

$E(X) = 0.383$

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$P(x = 4) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}}$

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$P(x = 4) = \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}}$

$P(x = 4) = \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}}$

$$ P(x = 4) = \frac{70 \times 84944276340}{35216131179263320}$

$P(x = 4) = 0.000169$

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

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3 years ago

What shape are the sides of a rectangular pyramid?

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Answer: Its a angle ...

Step-by-step explanation: Hope its fine.

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3 years ago

Please help for brainliest

Ilya [14]

Answer:

D.No solution

Step-by-step explanation:

These two function are parallel (same slope) , so they would never intersect

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3 years ago